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labwork [276]
1 year ago
9

6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been

Engineering
1 answer:
Katyanochek1 [597]1 year ago
5 0

During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

<h3>What is the pressure?</h3>

Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

Read more about the pressure :

brainly.com/question/25736513

#SPJ1

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One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal
7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

 

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m  

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 )  = ( (4h/kD)^1/2 )  

m =  ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04    

now lets find the value of h/mk    

h/mk = 10⁵ / ( 942809.04 × 30) =  0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb =  (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ 1 ]  / [cosh mL+  (h/mk) sinh mL]

so we substitute

0.893 =  [ 1 ]  / [cosh (942809.04 × L) +  (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

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4 0
3 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
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Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

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Answer:

The diamond ore will break and you won't get any diamonds.

Explanation:

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